TASC TCE 2024-style practice: Planes _1: x + 2y - 3z = 4 and _2: 2x + 4y + cz = d. (a) Show P(3, 2, 1) lies on _1 and Q(c, d4, -2) lies on _2. (b) For what value of c are the planes parallel? (c) With this c and d = 8, prove Q also lies on _1.

(a) For P on _1: 3 + 2(2) - 3(1) = 3 + 4 - 3 = 4, true. For Q on _2: 2c + 4(d4) + c(-2) = 2c + d - 2c = d, true. (2 marks) (b) Two planes are parallel when their normals are scalar multiples. The normal of _1 is (1, 2, -3) and the normal of _2 is (2, 4, c). Matching (2, 4, c) = 2(1, 2, -3) gives c = -6. (1 mark) (c) With c = -6 and d = 8, Q = (-6, 2, -2). Substitute into _1: (-6) + 2(2) - 3(-2) = -6 + 4 + 6 = 4, which equals the right side, so Q lies on _1. (2 marks)

§-Syllabus dot point

TASSpecialist MathematicsSyllabus dot point

How do we describe position, direction, lines and planes in three-dimensional space using vectors?

Use 3D vectors with dot and cross products to find angles, projections, lines and planes.

Three-dimensional vectors: magnitude, dot and cross products, angles, scalar projection, and equations of lines and planes for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

In three dimensions a vector has three components, $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ , often written as a column $(a_1, a_2, a_3)$ . The basic operations carry over from two dimensions, but two new ideas become essential: the cross product, and the description of lines and planes.

Magnitude and unit vectors

The magnitude (length) is

|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.

A unit vector in the direction of $\mathbf{a}$ is $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$ . Unit vectors are useful whenever you only care about direction, for example when writing the direction of a line.

The dot (scalar) product

\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}|\,|\mathbf{b}|\cos\theta,

where $\theta$ is the angle between the vectors. Rearranging gives the angle:

\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}.

Two non-zero vectors are perpendicular exactly when $\mathbf{a}\cdot\mathbf{b} = 0$ . The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$ , and the vector projection is $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}$ .

The cross (vector) product

The cross product produces a vector perpendicular to both inputs:

\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\mathbf{i} - (a_1 b_3 - a_3 b_1)\mathbf{j} + (a_1 b_2 - a_2 b_1)\mathbf{k}.

Its magnitude is $|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}|\,|\mathbf{b}|\sin\theta$ , which equals the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$ . Note that $\mathbf{a}\times\mathbf{b} = -(\mathbf{b}\times\mathbf{a})$ , so order matters.

Lines and planes

The vector equation of a line through point $\mathbf{r}_0$ with direction $\mathbf{d}$ is

\mathbf{r} = \mathbf{r}_0 + t\mathbf{d}, \qquad t \in \mathbb{R}.

The equation of a plane with normal $\mathbf{n} = (a,b,c)$ through point $\mathbf{r}_0$ is $\mathbf{n}\cdot(\mathbf{r} - \mathbf{r}_0) = 0$ , which expands to the Cartesian form $ax + by + cz = d$ where $d = \mathbf{n}\cdot\mathbf{r}_0$ . To find a normal to a plane through three points, take the cross product of two vectors lying in the plane.

A plane through three points

Find the Cartesian equation of the plane through $A(1,0,2)$ , $B(3,1,1)$ and $C(2,2,0)$ .

Step 1: Form two vectors lying in the plane

A plane is determined by a point and a normal. To get the normal we need two independent vectors that lie in the plane. We compute these by subtracting position vectors of the given points:

\overrightarrow{AB} = B - A = (3-1,\; 1-0,\; 1-2) = (2,1,-1), \qquad \overrightarrow{AC} = C - A = (1,2,-2).

Step 2: Compute the normal via the cross product

The cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ gives a vector perpendicular to both, hence perpendicular to the plane:

\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \big( (1)(-2)-(-1)(2),\; -[(2)(-2)-(-1)(1)],\; (2)(2)-(1)(1) \big).

Component by component: the $\mathbf{i}$ term is $-2 + 2 = 0$ ; the $\mathbf{j}$ term is $-(-4 + 1) = 3$ ; the $\mathbf{k}$ term is $4 - 1 = 3$ . So $\mathbf{n} = (0, 3, 3)$ , which we simplify to $(0,1,1)$ .

Step 3: Write the Cartesian equation using one known point

Substituting point $A(1,0,2)$ into $\mathbf{n}\cdot\mathbf{r} = d$ gives the constant:

d = \mathbf{n}\cdot\mathbf{r}_0 = (0)(1) + (1)(0) + (1)(2) = 2.

The plane is therefore:

y + z = 2.

Check with $B$ : $1 + 1 = 2$ . Check with $C$ : $2 + 0 = 2$ . Both satisfy it, confirming the equation is correct.

Final answer: $y + z = 2$ .

Putting it together

Many exam questions chain these tools: find a direction with subtraction, a normal with a cross product, an angle with a dot product, then assemble a line or plane equation. Lay your working out in clear named steps so each result is easy to reuse later in the question.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20236 marksThe point

P(1, -2, 3)

lies on the plane

\Pi: 3x + 4y - 2z = d

. (a) Determine

d

. (b) Write the line

L

through

P

parallel to direction

(2, \mu, -1)

in parametric form. (c) If

L

lies in

\Pi

, determine

\mu

Show worked answer →

(a) Substitute $P$ into the plane equation: $3(1) + 4(-2) - 2(3) = 3 - 8 - 6 = -11$ . So $d = -11$ . (2 marks)

(b) A line through $P$ with direction $(2, \mu, -1)$ is $(x, y, z) = (1, -2, 3) + t(2, \mu, -1)$ , that is $x = 1 + 2t$ , $y = -2 + \mu t$ , $z = 3 - t$ . (2 marks)

(c) If $L$ lies in the plane, its direction must be perpendicular to the plane's normal $(3, 4, -2)$ . Set the dot product to zero: $3(2) + 4\mu + (-2)(-1) = 6 + 4\mu + 2 = 0$ , so $4\mu = -8$ and $\mu = -2$ . Because $P$ already lies on the plane, this condition is sufficient. (2 marks)

TCE 20245 marksPlanes

\Pi_1: x + 2y - 3z = 4

and

\Pi_2: 2x + 4y + cz = d

. (a) Show

P(3, 2, 1)

lies on

\Pi_1

and

Q\left(c, \dfrac{d}{4}, -2\right)

lies on

\Pi_2

. (b) For what value of

c

are the planes parallel? (c) With this

c

and

d = 8

, prove

Q

also lies on

\Pi_1

Show worked answer →

(a) For $P$ on $\Pi_1$ : $3 + 2(2) - 3(1) = 3 + 4 - 3 = 4$ , true. For $Q$ on $\Pi_2$ : $2c + 4\left(\dfrac{d}{4}\right) + c(-2) = 2c + d - 2c = d$ , true. (2 marks)

(b) Two planes are parallel when their normals are scalar multiples. The normal of $\Pi_1$ is $(1, 2, -3)$ and the normal of $\Pi_2$ is $(2, 4, c)$ . Matching $(2, 4, c) = 2(1, 2, -3)$ gives $c = -6$ . (1 mark)

(c) With $c = -6$ and $d = 8$ , $Q = (-6, 2, -2)$ . Substitute into $\Pi_1$ : $(-6) + 2(2) - 3(-2) = -6 + 4 + 6 = 4$ , which equals the right side, so $Q$ lies on $\Pi_1$ . (2 marks)

TCE 20216 marks

Q

is the plane

4x - 3y + 6z = 36

. (a) Show

A(3, 2, 5)

lies on

Q

. (b) Determine the equation of the line joining

A

B(1, 2, -3)

. (c) Find the equation of the plane parallel to

Q

that contains

B

Show worked answer →

(a) Substitute $A$ : $4(3) - 3(2) + 6(5) = 12 - 6 + 30 = 36$ , which matches, so $A$ lies on $Q$ . (1 mark)

(b) The direction $\overrightarrow{AB} = B - A = (1 - 3, 2 - 2, -3 - 5) = (-2, 0, -8)$ , which simplifies to $(1, 0, 4)$ . The line is $(x, y, z) = (3, 2, 5) + t(1, 0, 4)$ , that is $x = 3 + t$ , $y = 2$ , $z = 5 + 4t$ . (3 marks)

(c) A parallel plane has the same normal $(4, -3, 6)$ , so it has the form $4x - 3y + 6z = k$ . Substitute $B(1, 2, -3)$ : $4 - 6 - 18 = -20$ . The plane is $4x - 3y + 6z = -20$ . (2 marks)